Question

5cosxctgx-5ctgx+2sinx=0

Answer 1

$5\, cosx\cdot ctgx-5\, ctgx+2sinx=0\; \; ,\; \; \; ODZ:\; x\ne \pi n,\; n\in Z\\\\5\, cosx\cdot \frac{cosx}{sinx}-5\cdot \frac{cosx}{sinx}+2sinx=0\\\\\frac{5\, cos^2x-5\, cosx+2sin^2x}{sinx}=0\; \; ,\; \; \frac{5\, cos^2x-5\, cosx+2(1-cos^2x)}{sinx}=0\; ,\\\\\frac{3\, cos^2x-5cosx+2}{sinx}=0\; \; \Rightarrow \; \; \left \{ {{3\, cos^2x-5\, cosx+2=0} \atop {x\ne \pi n,\; n\in Z}} \right. \\\\3\, cos^2x-5\, cosx+2=0\; ,\\\\t=cosx\; ,\; \; -1\leq t\leq 1\; \; ,\; \; 3t^2-5t+2=0\; ,\; \; D=1\; ,\; t_1=\frac{2}{3}\; ,\; t_2=1$

$a)\; \; cosx=\frac{2}{3}\; ,\; \; x=\pm arccos\frac{2}{3}+2\pi m\; ,\; m\in Z\\\\\left \{ {{x=\pm arccos \frac{2}{3}+2\pi m\; ,\; m\in Z} \atop {x\ne \pi n\; ,\; n\in Z}} \right. \; \; \Rightarrow \; \; x=\pm arccos \frac{2}{3}+2\pi m\; ,\; m\in Z\\\\b)\; \; cosx=1\; ,\; \; x=2\pi k\; ,\; k\in Z\; \; \Rightarrow \; \; \left \{ {{x=2\pi k\; ,\; k\in Z} \atop {x\ne \pi n\; ,\; n\in Z}} \right. \; \; \Rightarrow \; \; x\in \varnothing \\\\Otvet:\; \; x=\pm arccos \frac{2}{3}+2\pi m\; ,\; m\in Z\; .$